Tangents are drawn to the parabola y2 = 2x from the point Q (–2, 0). A and B be the point of contact of tangents above and below the x-axis respectively. QA & QB cuts the y-axis at D & C respectively.
(i) The area of Δ QCD (in sq. units):
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Ans.
(i)
Sol.
To find area ( Δ QCD) we have to find coordinates of A, B, C, D Let the equation of QA be y = mx + 
(Standard form)
Here a = 
(by y 2 = 4 
)
∴ y = mx + 
it passes through Q (–2, 0)
So 0 = – 2m + 
⇒ m = ± 
Point of contact 
⇒ (2, ± 2)
∴ A (2, 2), B (2, –2)
∴ Equation of QA is y = 
(x + 2) by putting x = 0
we get y = 1
So D (0, 1) Similarly by symmetry C (0, –1)
Now area of Δ QCD = 
× CD × OQ
(O : vertex of parabola) = 
× 2 × 2 = 2 sq. units
(ii)
Sol. Area of parallelogram CDAB
= 
× (AB + CD) × OP = 
× 6 × 2 = 6
(iii)
Sol. Area (QAB) = Area ( Δ QCD)
+ Area (Parallelogram CDAB)
= 2 + 6 = 8 sq. units
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